# Total Elastic Potential Energy 2

1. Jan 27, 2005

### kingyof2thejring

I need to know how to go about finding the speed.

A block B of mass 5 kg is fastened to one end of each of two springs. The other ends of the springs are attached to fixed points A and C, 4 metres apart on a smooth horizontal surface, as shown in the diagram.

Spring AB has natural length 2 metres and modulus of elasticity 30 N, while BC has natural length 1 metre and modulus 40 N (you may assume that the springs meet at the centre of B).

If the block is moved 0.5 metres towards C from its equilibrium position and then released, determine its speed as it passes through its equilibrium position.

this is what i have tried

Energy in string when moved 0.5m = total mech energy before the 0.5 movement
E= 30*(0.5)^2/(2*2) = 1.575 J

KE= 1/2*5*v^2 = 5/2*v^2 J and
E in string 30*(3/8)^2/2*2 = 7.93 J

7.93 + 5/2v^2 = 1.875
this leads me to taking square root of a -ve number

could someone help me!

2. Jan 28, 2005

### daster

I'm going to assume this is a follow-up to the question I did https://www.physicsforums.com/showthread.php?t=61316".

The springs already had an extension at the equilibrium position, so when you move it box towards C, the extension will change.
The extension in AB (x) is now: 0.5 + 8/11 = 27/22
The extension in BC (y) is now: 3/11 - 0.5 = -5/22 (i.e. compression)

Calculate the total mechanical energy at that point (which is only EPE). This should equal the total mechanical energy at the equilibrium position (which is KE & EPE).

It's important that you calculate the total energy. You shouldn't get the square root of a negative number.

Last edited by a moderator: Apr 21, 2017
3. Jan 28, 2005

### kingyof2thejring

cheers i can dig that